Physics Test - 41

Given,

Emf of the cell, \({E}_{1}=1.5 {V}\)

Balance point of the potentiometer, \(l_{1}=42 {~cm}\)

The cell is replaced by another cell of emf \({E}_{2}\).

New balance point of the potentiometer, \(l_{2}=63 {~cm}\)

The balance condition is given by the relation, 

\(\frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}}\)

\({E}_{2}={E}_{1} \times \frac{{l}_{2}}{l_{1}}=1.5 \times \frac{63}{42}=2.25 {~V}\)

Therefore, emf of the second cell is \(2.25 {~V}\).

As we know,

Work done by a variable force is given by,

\(W=\int_{x_{1}}^{x_{2}} F(x) d x\)

Work done, \((W)=\int_{0}^{t} F . d x=\int_{0}^{t} m a . d x\)

We know that, acceleration, \(a=\frac{d v}{d t}\)

\(\therefore W=\int_{0}^{t} m \frac{d v}{d t} \cdot d x \)

\(\Rightarrow W=\int_{0}^{t} m \frac{d(k \sqrt{x})}{d t} \cdot d x \)

\(\Rightarrow W=\int_{0}^{t} m \frac{k}{2 \sqrt{x}} \frac{d x}{d t} \cdot d x \)

\(\Rightarrow W=\int_{0}^{t} m \frac{k}{2 \sqrt{x}} v \cdot d x\)

\(\Rightarrow W=\int_{0}^{t} m \frac{k}{2 \sqrt{x}} k \sqrt{x} . d x \)

\(\Rightarrow W=\int_{0}^{t} m \frac{k^{2}}{2} \cdot d x \)

\(\Rightarrow W=m \frac{k^{2}}{2}[x]_{t=0}^{t=t}\) ...(1)

Given,

Velocity \(v=k\sqrt{x}\)

As we know,

Velocity, \(v=\frac{d x}{d t}\)

\(\therefore k \sqrt{x}=\frac{d x}{d t} \)

\(\Rightarrow k d t=\frac{d x}{\sqrt{x}}\)

On integrating, we get

\(kt =2 \sqrt{ x }\)

At \(t=0, x=0\)

\(\therefore x=\frac{k^{2} t^{2}}{4}\)...(2)

Substituting (2) in (1), we get

\(W=m \frac{k^{2}}{2}\left[\frac{k^{2} t^{2}}{4}\right]\)

\(\Rightarrow W=\frac{m k^{4} t^{2}}{8}\)

In series combination \(\Delta l=l_1+l_2\)

\( Y=\frac{\frac{F }{ A}}{\frac{\Delta l }{ l}} \Rightarrow \Delta l=\frac{F l}{A Y} \)

\( \Rightarrow \Delta l \propto \frac{l}{Y}\)

Equivalent length of rod after joing is \(=2 l\)

As, lengths are same and force is also same in series

\( \Delta l=\Delta l_1+\Delta l_2 \)

\( \frac{l_{c q}}{Y_{c q}}=\frac{l}{Y_1}+\frac{l}{Y_2} \Rightarrow \frac{2 l}{Y}=\frac{l}{Y_1}+\frac{l}{Y_2} \)

\( \therefore Y=\frac{2 Y_1 Y_2}{Y_1+Y_2}\)

Unit of magnetic flux is weber.

Magnetic flux is a physical quantity that measures the absolute magnitude of a magnetic field passing through a plane (such as a coil of a conducting wire). It is denoted by abbreviation . Its SI unit is Weber.

Since magnetic flux \((\varphi)=\mathrm{B} \mathrm{A}\)

Where, \(\mathrm{B}\) = megnatic field and \(\mathrm{A}\) = area

The \(\mathrm{SI}\) unit of magnetic flux \(=\mathrm{SI}\) unit of magnetic field × \( \mathrm{SI}\) unit of area \(=\) tesla meter\(^{2}=\mathrm{T} \mathrm{m}^{2}\)

Since, 1 Weber \(=1 \mathrm{~T} \mathrm{m}^{2}\)

Thus the SI unit of magnetic flux is \(\mathrm{T} \mathrm{m}^{2}\) and which is equal to weber \((\mathrm{Wb})\).

Considering the above figure let us say a swimmer at point \(B\) wishes to cross the river such that the time taken is the minimum. Therefore he has to make sure he is moving in a perpendicular path i.e. in the above figure is \(\mathrm{AB}\). The direction of the flow of water is \(BD\) with speed \(\left(V_{W}\right)\). Hence the swimmer actually has to swim with his speed \(\left(V_{S}\right)\) that is still water in direction \(CB\) such that he actually moves perpendicular. It is given to us that the average speed of the swimmer in the river is \(V_{A}\) i.e. along with path \(AB\). Applying the Pythagoras theorem to the triangle \(ABD\) 

we get, \(V_{S}^{2}=V_{A}^{2}+V_{W}^{2}\)

\(\Rightarrow 5^{2}=3^{3}+V_{W}^{2}\) 

\(\Rightarrow V_{W}^{2}=25-9\) \(=16\) 

\(\Rightarrow V_{W}=4\)

The de-Broglie wavelength of a particle of mass \(\mathrm{m}\) and moving with velocity \(\mathrm{v}\) is given by,

\(\lambda=\frac{\mathrm{h}}{\mathrm{mv}} \quad(\because \mathrm{p}=\mathrm{mv})\)

de-Broglie wavelength of a proton of mass \(\mathrm{m}_{1}\) and kinetic energy \(\mathrm{k}\) is given by,

\(\lambda_{1}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{1} \mathrm{k}}} \quad(\because \mathrm{p}=\sqrt{2 \mathrm{mk}})\)

\(\Rightarrow \lambda_{1}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{1} \mathrm{qV}}} \ldots . .\) (i) \(\quad[\because \mathrm{k}=\mathrm{qV}]\)

For an alpha particle mass \(\mathrm{m}_{2}\) carrying charge \(\mathrm{q}_{0}\) is accelerated through potential \(\mathrm{V}\), then,

\(\lambda_{2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{2} \mathrm{q}_{0} \mathrm{~V}}}\)

\(\because\) For \(\alpha-\) particle \(\left({ }_{2}^{4} \mathrm{He}\right): \)

\(\mathrm{q}_{0}=2 \mathrm{q}\) and \(\mathrm{m}_{2}=4 \mathrm{~m}_{1}\)

\(\therefore \lambda_{2}=\frac{\mathrm{h}}{\sqrt{2 \times 4 \mathrm{~m}_{1} \times 2 \mathrm{q} \times \mathrm{V}}}.....\) (ii)

The ratio of corresponding wavelength, from Eqs. (i) and (ii), we get,

\(\frac{\lambda_{1}}{\lambda_{2}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{1} \mathrm{q} \mathrm{V}}} \times \frac{\sqrt{2 \times \mathrm{m}_{1} \times 4 \times 2 \mathrm{q} \mathrm{V}}}{\mathrm{h}}\)

\(=\frac{4}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\)

\(\Rightarrow \frac{\lambda_{1}}{\lambda_{2}}=2 \sqrt{2}\)

An ion is an atom or a group of atoms that does not equal the number of electrons to the number of protons. Electrons have a negative charge, whereas there is a positive charge for protons. This results in a negative charge when an atom gains electrons. An anion is called this form of ion. This results in a positive charge when an atom loses electrons. A cation is considered a positively-charged ion. Usually, positive ions are metals or behave like metals. Just like atoms may lose electrons to become cations, others can absorb electrons and become anions that are negatively charged.

Tension due to a string is found using the free body diagram,

We know thattime period ofsimple pendulum is:

\(T=2 \pi \sqrt{\frac{l}{g}}\)

Taking log in both sides:

\(\log T=\log 2+\log \pi+\frac{1}{2} \log l-\frac{1}{2} \log g\)

Differentiating it we get

\(\frac{d T}{T}=\frac{1}{2} \frac{d l}{l}-\frac{1}{2} \frac{d g}{g}=-\frac{1}{2} \frac{d g}{g}(\therefore l\) is constant \()\)

\(\%\) change in time period

\(=\frac{d T}{T} \times 100=-\frac{1}{2} \frac{d g}{g} \times 100 \)

\(=-\frac{1}{2}\left(\frac{-2}{100}\right) \times 100\)

\(=1 \% \text { (increase) }\)